1. 15 Lot-by-Lot Acceptance Sampling for Attributes Chapter 15
Dr. Shokri Selim, KFUPM

2. The acceptance sampling problem
An old method used in the 1930’s and 40’s.
The purpose of AS is to inspect received lots of products and decide whether to accept or reject the lot; lot disposition, or lot sentencing.
Accepted lots are put into production
Rejected lots may be returned to supplier or subjected to other lot-disposition action
Sampling methods may also be used during various stages of production
Chapter 15
Dr. Shokri Selim, KFUPM

3. The purpose of AS is sentencing lots and not estimate lot quality.
Note that:
The purpose of AS is sentencing lots and not estimate
lot quality.
2. It could happen that, lots of same quality be sentenced
differently based on the sample.
3. AS is an audit tool
4. Control charts are used to signal departures from
quality.
Chapter 15
Dr. Shokri Selim, KFUPM

4. Approaches to lot sentencing
Accept without any inspection
Supplier process is very good and defectives are rare or
there is no economic justification to look for defectives
100% inspection
Use if defective components can cause high failure cost or
supplier process does not meet specifications
Acceptance sampling
Chapter 15
Dr. Shokri Selim, KFUPM

5. When is AS used?
Chapter 15
Dr. Shokri Selim, KFUPM

6. Advantages of sampling
Chapter 15
Dr. Shokri Selim, KFUPM

7. Disadvantages of sampling
There is a risk of accepting “bad” lots and rejecting “good” lots.
Less information is generated about the product or about the process that manufactured it.
Acceptance sampling requires planning and documentation where as 100% inspection does not
Chapter 15
Dr. Shokri Selim, KFUPM

8. Types of sampling plans
One major classification is by data type,
variables and attributes
Another is based on the number of samples
required for a decision.
Chapter 15
Dr. Shokri Selim, KFUPM

9. Types based on number of samples
Single-sampling plans
Select a sample of size n. If the number of defectives ≤ c, accept lot, else, reject the lot.
Double-sampling plans
Select a sample of size n, then depending on the number of defective
accept the lot
reject the lot
take a second sample and decide on both samples
Multiple-sampling plans
similar to double but with more than 2 samples
Sequential-sampling plans
units are selected one at a time and decision to accept or reject or continue is made
Chapter 15
Dr. Shokri Selim, KFUPM

10. Factors to consider include: Administrative efficiency
Single-, double-, multiple-, and sequential sampling plans can be designed to produce equivalent results.
A lot of the some quality level has the same probability of being accepted by these plans.
Factors to consider include:
Administrative efficiency
Type of information produced by the plan
Average amount of inspection required by plan
Impact of the procedure on manufacturing flow
Chapter 15
Dr. Shokri Selim, KFUPM

11. Lot formation
There are a number of important considerations in forming lots for inspection, including:
Lots should be homogeneous.
Same machine, same raw material, same operator
Larger lots are preferred over smaller ones.
More economic
Lots should be conformable to materials-handling systems used in both supplier and consumer facilities.
Lot packaging minimizes risk of damage
Selection of sample is easier
Chapter 15
Dr. Shokri Selim, KFUPM

12. Random sampling
Assign a number to each item and select n random numbers to form a sample or
Randomly select the length, depth and width in the container or
Stratify the lot into layers, then cubes.
Chapter 15
Dr. Shokri Selim, KFUPM

13. Guidelines for using acceptance sampling
The selection of a sampling plan depends on the objective and the history of the supplier.
Non-static nature of AS plans
If supplier is known for quality, start with sampling plan for attributes. If quality is proven, may use skip-a-lot policy. If capability is high may stop sampling.
If supplier quality is not known, use an attribute plan. If quality is good may use a variable plan, and help them in SPC
Companies start with AS and shift to SPC as the quality improves.
Chapter 15
Dr. Shokri Selim, KFUPM

14. Single sampling plans for attributes
Definition of a single sampling plan
The plan is defined by the sample size n and
the acceptance number c.
If the number of defectives, d ≤ c accept the lot,
else reject the lot.
Chapter 15
Dr. Shokri Selim, KFUPM

15. Types of Lot Size Type A: Lot size is finite
Type B: Lot size in infinite
Chapter 15
Dr. Shokri Selim, KFUPM

16. Type B OC curves
The OC curve gives the probability of accepting the lot given the lot fraction defective
Main assumption: lot size is very large
Let p = probability a unit is defective
d = number of defectives in a sample of size n
Probability of accepting a lot:
Chapter 15
Dr. Shokri Selim, KFUPM

17. Sample OC curve
Chapter 15
Dr. Shokri Selim, KFUPM

18. The ideal OC curve Reject lot if p > 0.025
Can we choose n and c that give similar OCC?
Chapter 15
Dr. Shokri Selim, KFUPM

19. Effect of n
Chapter 15
Dr. Shokri Selim, KFUPM

20. Effect of C
Chapter 15
Dr. Shokri Selim, KFUPM

21. Effect of n and c on OC curves
Chapter 15

22. Type-A and Type-B OC curves
If the lot size is vey large, we use the binomial
distribution to model the P(d ≤ c)
If the lot size is finite we use the hypergeometric
distribution.
Chapter 15
Dr. Shokri Selim, KFUPM

23. Constructing Type A OCC
Let
N be the lot size
D be the number of defectives in the lot
N be the sample size
C be the maximum number of defectives allowed
DEMO
Chapter 15
Dr. Shokri Selim, KFUPM

24. Relation between Types A and B
If N is large (N ≥ 10n ) both graphs are close.
DEMO
Type A and Type B OC curves
Chapter 15
Dr. Shokri Selim, KFUPM

25. Behavior of Type B OC curve for c = 0
Plans with c = 0.
OC curve far from the ideal OCC.
Pa falls sharply as p increases
Chapter 15
Dr. Shokri Selim, KFUPM

26. Behavior of Type A OC curve for c = 0
Plans with c = 0 and N = 10n
OC curve far from the ideal OCC.
Pa falls sharply as p increases
N = 10n but the OCCs behave differently.
p = D/N
Chapter 15
Dr. Shokri Selim, KFUPM

27. Specific points on the OC curve
Acceptable quality level, AQL = the least quality
level for the supplier’s process that a consumer
would consider to be acceptable. The consumer
assigns high acceptance probability to it.
Lot tolerance percent defective, LTPD =
Rejectable quality level, RQL =
the least quality level, that the consumer is willing to accept with small acceptance probability.
We can design a plan that almost satisfies both conditions.
Chapter 15
Dr. Shokri Selim, KFUPM

28. Designing a single-sampling plan with a specified OC curve
p1 = AQL
p2 = RQL
DEMO
Chapter 15
Dr. Shokri Selim, KFUPM

29. N ≥ 10 n
Binomial monograph
Chapter 15
Dr. Shokri Selim, KFUPM

30. N ≥ 10 n
Binomial monograph
Chapter 15
Dr. Shokri Selim, KFUPM

31. Chapter 15

32. Rectifying inspection
Incoming lots
Fraction
Defectives P0
Outgoing lots
Fraction
Defectives
P1 < P0
Inspection
activity
Rejected lot has 0 defectives
Accepted lot has P0 fraction defectives
No of defectives = P0(N-n)
Average number of defective units =
pa (N – n)p0 + ( 1 – pa)x0
Average outgoing quality, AOQ = pa(N – n)p0/N
Chapter 15
Dr. Shokri Selim, KFUPM

33. Example
N = 1000, n =100, c = 3, p = 0.01
pa =
Average outgoing quality =
AOQ = pa(N – n)p/N = 0.009
Chapter 15
Dr. Shokri Selim, KFUPM

34. AOQL is the maximum point on the curve
AOQ for N very large
AOQ = pa(N – n)p/N
= pa(1 – n/N)p
≈ pa p
AOQL is the maximum point on the curve
AOQ limit = worst AQL
For the example; AOQL =
Chapter 15
Dr. Shokri Selim, KFUPM

35. Average total inspection
Number of inspected items: If d ≤ c, n units will be inspected. If d > c, N units will be inspected. ATI = pa n + ( 1 – pa )*N = n + ( 1 – pa )*( N – n) If N= 1000, n= 100, c = 3, p= 0.01 pa = and ATI =
Chapter 15
Dr. Shokri Selim, KFUPM

36. ATI = n + ( 1- pa )*( N – n)
Does the behavior of the graph makes sense?
ATI for sampling plan: n = 89, c = 2
for lot sizes of 1000, 5000, 10,000
Chapter 15
Dr. Shokri Selim, KFUPM

37. Selection of n and c based on AOQL and ATI
If AOQL is specified, the solution is not unique
We find n and c that minimize ATI given some AOQL value
Chapter 15
Dr. Shokri Selim, KFUPM

38. Double, multiple, and sequential sampling
Double Sampling Plans
n1 = sample size on the first sample
c1 = acceptance number of the first sample
n2 = sample size on the second sample
c2 = acceptance number of the second sample
If d1 in the first sample is ≤ c1 accept the lot
If d1 in the first sample is > c2 reject the lot
Otherwise take 2nd sample.
If d1 + d2 ≤ c2 accept the lot
Otherwise, reject the lot.
Chapter 15
Dr. Shokri Selim, KFUPM

39. Chapter 15
Dr. Shokri Selim, KFUPM

40. Chapter 15
Dr. Shokri Selim, KFUPM

41. Advantage of double sampling over single sampling plans
HW: read the advantages and disadvantages
Chapter 15
Dr. Shokri Selim, KFUPM

42. Constructing the OC curve for double sampling plans
Chapter 15
Dr. Shokri Selim, KFUPM

43. Supplementary OC curve Primary OC curve Supplementary OC curve
Chapter 15
Dr. Shokri Selim, KFUPM

44. Example n1 = 50, C1 = 3, n2 = 150, C2 = 6 Chapter 15
Dr. Shokri Selim, KFUPM

45. The average sample number
It is the average number of inspected units
ASN = n1 + n2 Pr( c1 < d1 ≤ c2 )
What is the assumption here ?
The assumption
We complete the inspection of the second sample even after the total number of defectives has exceeded c2
Chapter 15
Dr. Shokri Selim, KFUPM

46. Curtailment
It is stoppage of sampling when the rejection condition is satisfied
Do not do it with the first sample
Why ?
To be able to estimate the fraction defective.
However, can do on the second sample.
Chapter 15
Dr. Shokri Selim, KFUPM

47. Rectifying Inspection with double sampling
If all defective items are discovered, either in sampling or 100% inspection, and are replaced with good ones:
Chapter 15
Dr. Shokri Selim, KFUPM

48. Multiple Sampling Plans
Example:
Cumulative
sample size
Acceptance
number
Rejection 20 3 40 1 4 60 5 80 7
100 8 9
Chapter 15
Dr. Shokri Selim, KFUPM

49. At the completion of stage i:
Cumulative
sample size
Acceptance
number
Rejection 20 3 40 1 4 60 5 80 7
100 8 9
At the completion of stage i:
If d1 + d2 + … + di ≤ acceptance number → Accept lot
If d1 + d2 + … + di ≥ rejection number → Reject lot
Otherwise take the next sample
Usually, the first sample is inspected 100%.
Usually, subsequent samples are subject to curtailment
Chapter 15
Dr. Shokri Selim, KFUPM

50. Take 2nd sample, what value of d2 will result in accepting lot?
Cumulative
sample size
Acceptance
number
Rejection 20 3 40 1 4 60 5 80 7
100 8 9
Suppose d1 = 1
Take 2nd sample, what value of d2 will result in accepting lot?
what value of d2 will result in rejecting lot?
suppose d2 = 1
Take 3nd sample, what value of d3 will result in accepting lot?
what value of d3 will result in rejecting lot?
Chapter 15
Dr. Shokri Selim, KFUPM

51. The values of the d’s that lead to lot acceptance
Cumulative
sample size
Acceptance
number
Rejection 20 3 40 1 4 60 5 80 7
100 8 9 3 1 2 d1 4 d2 5 d3 7 d4 8 9 d5
Chapter 15
Dr. Shokri Selim, KFUPM

52. Item by Item Sequential Sampling Plans
The x-axis shows cumulative number of items inspected. The y-axis shows cumulative number of defectives If the point is between the two lines, take one more item If the point falls on or above the top line, reject lot If the point falls on or below the bottom line, accept lot
Chapter 15
Dr. Shokri Selim, KFUPM

53. The acceptance and rejection lines
Given p1 and 1-α, p2 and β.
Suppose p1 = 0.01, α = 0.05,
p2 = 0.06, β = 0.1
Chapter 15
Dr. Shokri Selim, KFUPM

54. Truncation of sampling
Sequential sampling could be truncated if the number of units inspected reaches three times the sample size of the equivalent single sample plan
For the previous example, the equivalent single sample plan has n = 89. If sentencing does not takes place after 267 units, stop sampling and accept lot
Chapter 15
Dr. Shokri Selim, KFUPM

55. Homework
Project 1:
Suppose the fraction defective is 0.05, find the ASN for double sampling plan defined by (n1 = 50, c1 = 1, n2 = 50, c2 = 2), in case of curtailment at the second sample
Project 2:
A single sample plan has n = 50; find c that will result in least ATI and AOQL ≤ 0.01
Project 3:
Consider a double sampling plan with n1=50, c1=1, c2 =2. Find the smallest n2 that will result in AQL = 0.01 with α ≤ 0.05, and RQL = 0.1 with β ≤ 0.2.
Chapter 15
Chapter 15
Dr. Shokri Selim, KFUPM
Dr. Shokri Selim, KFUPM 55